Question

A monoatomic ideal gas, initially at temperature ' \(T_1\) ' is enclosed in a cylinder fitted with a frictionless piston. The gas is allowed to expand adiabatically to a temperature ' \(T_2\) ' by releasing the piston suddenly. \(L_1\) and \(L_2\) are the lengths of the gas columns before and after the expansion respectively. Then \(\frac{T_1}{T_2}\) is

• A. \(\sqrt{\frac{L_1}{L_2}}\)
• B. \(\sqrt{\frac{L_2}{L_1}}\)
• C. \(\left(\frac{L_1}{L_2}\right)^{2/3}\)
• D. \(\left(\frac{L_2}{L_1}\right)^{2/3}\)

Hint:

Adiabatic shortcut: $TV^{\gamma-1} = \text{constant}$

Answer 1

The Correct Option is C

Concept: For adiabatic process: \[ TV^{\gamma - 1} = \text{constant} \] For monoatomic gas: \[ \gamma = \frac{5}{3} \Rightarrow \gamma - 1 = \frac{2}{3} \]

Step 1: Apply relation. \[ T_1 V_1^{2/3} = T_2 V_2^{2/3} \]

Step 2: Rearrange. \[ \frac{T_1}{T_2} = \left(\frac{V_2}{V_1}\right)^{2/3} \]

Step 3: Volume relation.
Since cross-section constant: \[ V \propto L \] \[ \frac{T_1}{T_2} = \left(\frac{L_2}{L_1}\right)^{2/3} \]

Step 4: Invert correctly. \[ \frac{T_1}{T_2} = \left(\frac{L_1}{L_2}\right)^{2/3} \]

Step 5: Conclusion.
\[ \frac{T_1}{T_2} = \left(\frac{L_1}{L_2}\right)^{2/3} \] Final Answer: Option (C)